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Limits, what they are and what they do

  1. May 24, 2003 #1
    My text book failed to address how to find the epsilon and the delta of a Limit. Can somebody explain to me what they are and what they do?
     
    Last edited by a moderator: Feb 5, 2013
  2. jcsd
  3. May 24, 2003 #2
    Welcome to the forums!!!!!

    What class is this for? I have no idea what those words mean! Except Delta...
     
  4. May 24, 2003 #3
    Calculus. Where else do you think they will teach Limits?
     
  5. May 24, 2003 #4
    I figured it'd be calc, but again I have no clue. I just finished precalc but that wasn't an issue there.

    I may take calc in 1 variable as an elective fill-in, but more likely I'll take something else more usefull in my career.
     
  6. May 25, 2003 #5

    Hurkyl

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    Limits don't have an epsilon (ε) and a delta (δ) per se...


    The idea of a limit is that it's supposed to capture our idea of "approaching" a number. For example, if I have the function:

    f(x) = (x2 - 4x + 3) / (x - 1)

    I know that f(1) does not exist (because 0/0 is undefined). However, if I compute some values of f near 1:

    f(1.1) = -1.9
    f(1.01) = -1.99
    f(1.001) = -1.999
    et cetera

    It appears to me that f(x) is approaching -2 as x approaches 1. Incidentally, we can write this in symbols as:

    f(x) -> -2 as x -> 1

    (where -> is supposed to be an arrow, not a dash followed by a greater than sign)

    So the question is, how would we prove this? How can we put into symbols the idea that as x approaches 1, f(x) approahces -2?

    Well, it's difficult to describe the idea of a moving point in mathematics, so instead lets just let x be some point near 1, and require that it has to be close to 1. I haven't decided just how close to 1 we should make x, so let's use a variable for this distance.

    Let the symbol δ be that variable. I shall require the distance between x and 1 to be less than δ. IOW:

    |x - 1| < &delta;

    Also, don't forget x isn't allowed to be exactly 1! Thus:

    0 < |x - 1| < &delta;

    Now, what about the value of f(x)? Let's use the same idea! Let's introduce the variable &epsilon; to be the bound on how close f(x) must be to -2. IOW, I want:

    |f(x) - (-2)| < &epsilon;

    Now, let us combine these two ideas. I want to say "When x is near 1, f(x) is near -2", so putting that into symbols:

    I want to prove that: 0 < |x - 1| < &delta; implies |f(x) - (-2)| < &epsilon;

    But we still have to figure out what &delta and &epsilon should be!

    Well, let's defer that a little bit longer and figure out how they should be related. Our end goal is to prove that f(x) is approaching (-2), so let's fix &epsilon; and see if we can find a formula for &delta; that makes the implication true.

    (note: These next steps are typical of so-called "&epsilon;-&delta;" proofs)

    Goal: Fnid &delta; so that:
    0 < |x - 1| < &delta; implies |f(x) - (-2)| < &epsilon;

    First, plug in what f(x) is:

    |(x2 - 4x + 3) / (x - 1) - (-2)| < &epsilon;

    Multiply through by |x-1|:

    |(x2 - 4x + 3) + 2(x-1)| < &epsilon; |x-1|
    |x2 - 2x + 1| < &epsilon; |x-1|
    |(x - 1)2| < &epsilon; |x-1|

    now divide

    |x-1| < &epsilon;

    And now it's clear how to choose our value for &delta;! If we simply choose &delta; to be equal to &epsilon;, then we're appear to be guaranteed that |x-1| < &epsilon;. However, the steps we did above are in the reverse order of what we need to prove this claim, so lets work forwards:

    Choose &delta; to be equal to &epsilon;
    0 < |x - 1| < &delta;
    substituting yields: |x - 1| < &epsilon;
    Multiply by |x-1|: |x2 - 2x + 1| < &epsilon; |x-1|
    Add and subtract inside the ||: |(x2 - 4x + 3) + 2(x-1)| < &epsilon; |x-1|
    Substitute f: |f(x) - (-2)| < &epsilon;

    So we've proven our goal! For any &epsilon; we choose, we may select &delta; to be equal to &epsilon; and then the following is true:

    0 < |x - 1| < &delta; implies |f(x) - (-2)| < &epsilon;

    IOW, no matter how close we want f(x) to be to (-2), we are able to sufficiently narrow the freedom on x so that f(x) is close enough to (-2) for all allowed values of x.


    But we still haven't chosen an actual value for &epsilon;! And this is the key to how the definition of a limit corresponds to our intuitive idea. No matter how small we want the error to be, we can restrict x to a small enough interval around 1 so that f(x) is always within our error bounds!.

    Anyways, that's the motivation for the whole thing. Summing it up in the rigorous definition:

    f(x) -> L as x -> a if and only if
    For every &epsilon;>0 there exists a &delta;>0 such that for all x:
    0 < |x - a| < &delta; implies |f(x) - L| < &epsilon;


    And to prove that a limit is a particular value using this definition, the typical strategy is what we did above; we "solve" for &delta; in terms of &epsilon; and then prove our solution yields the implication required by the definition.


    And getting back to my original statement, &delta; and &epsilon; are dummy variables; we could use other letters for them if we like... we just tend to use &delta; and &epsilon; by convention. These proofs are confusing to beginners, and it helps to have a consistent variable to eliminate a source of confusion.

     
  7. May 26, 2003 #6
    l'Hospital's rule!

    Hi Hurkyl,
    Also: f(.9) = -2.1
    f(.99) = -2.01
    f(.999) = -2.001 et cetera
    and I learned in college that the validation of this technique was referred to as "Cauchy Critereon" - "For every neighborhood, no matter how small, - - -etc etc."

    Of course there is no argument that the limit of the example is really - 2; and when I tried to quick fix using l'Hospital's treatment, your arduous but elegantly explicit post was verified post haste.
    The Rule states that if the ratio of two functions involves division by zero at some limiting value, the ratio of the derivatives of those functions (or 2nd,3rd etc derivatives) may indicate the existance of a limit and, as I later learned, the actual value of that limit.
    It's easy to see how it works for the limit (as x ->0) of [sin x]/x which is [cos x]dx/dx =1 Q.E.D.
    similarly for your example the derivative ratio is [2x -4]dx/dx which when evaluated at x=1 gives [2x1 - 4] = -2 Q.E.D. Cheers, Jim
     
  8. May 26, 2003 #7

    Hurkyl

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    I guess a followup is in order so as not to give Prudens the impression that calculus is horribly complicated.


    As NEO mentioned, there are other ways to compute limits. It is important to learn the definition of a limit so you understand what the mathematics is trying to say (and so you can get that problem right on the test), but for practical computation, different techniques are used, which you will learn in a few sections.
     
  9. May 27, 2003 #8
    Re: l'Hospital's rule!

    using the epsilon (£`) and a delta (£_) is to prove the limit is logic
    just like to prove limit x( x approach to 3 )= 3
    although you know the answer is 3
    but for the mathematics logic
    you should show the equation is logic
    and epsilon (£`) and a delta (£_) is the way
    just like what hurkyl show
    this are not using to find the answer of limet !!!
     
  10. May 27, 2003 #9
    What if a limit doesn't exist? Can we or should we use the epsilon-delta argument to show a limit doesn't exist? (I think the answer is yes and we just need to proof
    "For every &epsilon;>0 there exists a &delta;>0 such that for all x:
    0 < |x - a| < &delta; implies |f(x) - L| <&epsilon; " isn't true)
    Since we are working backward to find the values of &epsilon;
    and &delta;, I think it is harder to do the proof, right? Would you please give me an example using epsilon-delta argument to prove a limit doesn't exist? I've never met one in class.

    Thanks
     
  11. May 27, 2003 #10
    what you mean the limit does not exist??
    did you mean the limit go to infinity? or discontinuity??
    i think this can help....
    for the infinite large definition
    For each M>0 there exists a £_>0 such that
    0 < |x - a| < £_ implies f(x) >M
    infinite small definition is
    For each M>0 there exists a £_>0 such that
    0 < |x - a| < £_ implies f(x) <-M
    and continuity definition is
    For each £`>0 there exists a £_>0 such that
    0 < |x - a| < £_ then |f(x) - f(a)| <£`
    if the limit can not find some £_ in this definition
    that mean the function discontinuity at a
     
  12. May 27, 2003 #11

    wrong!! Hurkyl...you wrong!!
    using the epsilon (£`) and a delta (£_) is to prove the limit is logic
    just like to prove limit x( x approach to 3 )= 3
    although you know the answer is 3
    but for the mathematics logic
    you should show the equation is logic
    just like the example you write
    you already know the answer of the limit
    and you just prove the equation is logic
    l'Hospital's rule is use to find the answer of the limit
    this two skill are different.....
    epsilon (£`) and a delta (£_) can't do what l'Hospital rule do
    and L'Hospital rule can't do what epsilon (£`) and a delta (£_) do
    you are not grasp the meaning of epsilon (£`) and a delta (£_)
     
  13. May 27, 2003 #12

    Hurkyl

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    You use the &epsilon;-&delta; definition to prove various theorems about limits and then you use those theorems to actually compute the values of limits.

    For instance, for my example problem, we had

    f(x) = (x2 - 4x + 3) / (x - 1)

    However, notice the numerator can be factored into (x - 1) * (x - 3). If we let

    g(x) = x - 3

    then g(x) = f(x) for all x [x=] 1. We can prove, using the &epsilon;-&delta; definition the theorem that:

    If f(x) = g(x) for all x such that p < x < a and a < x < q (for some p and q)
    and f(x) -> L as x -> a
    then g(x) -> L as x -> a

    So then to find the limit of f(x) as x approaches 1, we simply need to find the limit of g(x) as x approaches 1.

    We can also prove the following theorem, again using the &epsilon;-&delta; definition:

    If h(x) is a polynomial, then h(x) -> h(a) as x -> a

    So then we know that g(x) -> g(1) as x -> 1.

    And we combine the two theorems to prove
    f(x) -> -2 as x -> 1
     
  14. May 27, 2003 #13
    no...no...
    first...you already know the answer of the g(x)
    second , if f(x) is unknown
    how can you assume f(x)--> a and g(x)--> a
    if f(x) is so complex
    how can you find some function g(x) is eqaul to f(x)
    you assume the f(x)--> a and g(x)--> a
    because you already know the answer of f(x)
    if not, you will not set the g(1)--> 2
    i told you already £`-£_ definition just to prove the limit is logic when you already know the answer
    L'Hospital rule is find the answer of the unknown limit
    but £`-£_ definition can't do that
     
  15. May 28, 2003 #14


    According to Hurkyl's last post, he didn't know the limit value of both f(x) and g(x) at the very beginning. He only knew f(x) can be simplified to g(x), where x [x=] 1, after factorization.

    He didn't set limit value of g(x) tends to 2 as x tends to 1.
    He used epsilon-delta (e-d) defination to prove the following :

    If f(x) = g(x) for all x such that p < x < a and a < x < q (for some p and q) and f(x) -> L as x -> a
    then g(x) -> L as x -> a ........(1)

    In his example, f(x)=g(x) for all x such that p<x<1 and 1<x<q for some p and q and f(x) -> L as x ->a. (We still don't know the the limit valye, L up to now)

    Then he proved, using e-d defination that
    If h(x) is a polynomial, then h(x) -> h(a) as x -> a .......... (2)

    By combining (1) and (2), we know that we can substitute x=1 into g(x) to get the limit value of f(x)

    We don't know the limit value of f(x) before step (1) and (2) but we've used e-d defination to show that we can use (1) and (2) to compute the limit value of f(x)

    I think the L'hospital rule is somehow based on the e-d defination, but not the other way round.
     
  16. May 28, 2003 #15
    ok
    now i give you a question
    then you try to use the £`-£_ definition to find the answer
    but....don't use the L'Hospital rule to find the answer
    ok??
    limit [(1/x)-1/((e^x)-1)]=?? ....(x-->0)
    if you can do this with £`-£_ definition
    show you step
    and.....don't use the other way to find the answer 1st!!!
     
  17. May 28, 2003 #16

    Hurkyl

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    I'm not sure what point you're trying to make by this.

    I could just use the &epsilon;-&delta method to prove L'Hopital's rule and use L'Hopital's rule to compute the limit, and I would be fully within the constraints of your challenge because I did the whole thing using the &epsilon;-&delta; definition.

    That's why we use things like L'Hopital's rule to evaluate limits in practice; they are rigorously proven in terms of the &epsilon;-&delta; definition, yet are simpler to apply... but since we've proven L'Hopital's rule in terms of the &epsilon;-&delta; definition, we can, in principle, just cut-paste the proof of L'Hopital's rule (and other theorems) into our proofs to give a proof that uses nothing but &epsilon;-&delta;, though it could be very long and complex. We use theorems in mathematics so we don't have to cut and paste.
     
    Last edited: May 28, 2003
  18. May 28, 2003 #17
    just like what i say before
    £`-£_ definition can't use to find the answer of limit
    this definition just use to proof your answer is right
    but L'Hospital just only can find the answer
    this 2 skill is different
    but your people say £`-£_ is "the other ways to compute limits"
    this is absolutely wrong!!
     
  19. May 28, 2003 #18

    Hurkyl

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    Use of L'Hopital's rule is most certainly proof that the limit has a particular value.
     
  20. May 28, 2003 #19
    i am lazy to say again
    you really don't know what differeent with "prove it exist" and "find the value"
     
  21. May 29, 2003 #20
    Unique evaluation is corollary of existence

    Hi again Hurkyl,
    When I check the profiles of this string, I see that posters are students (high-school and Bacc)and because I learned about the power of l'Hospital's rule and Cauchy Criterion in graduate studies, I submit that the adversarial nature of this string is likely caused by the absence of a complete deck. Challenge is often the route to learning that must be tolerated and that is okay.
    Clearly here, some of these students have possibly learned something new (the rule and the criterion) from our postings and I propose that they check them out at the school library or Google.
    I have discovered that if the derivative of a function divides the function itself, the result is the Newton-Raphson root-extraction method (see Greg's other forum).
    newton1 will have trouble when he tries to understand how some stepwise functions have dual limits dependent on whether analysis is from the left or the right; the "proof of the pudding is - -" and all that rot! Cheers, Jim PS: Thanks for your patience with these kids.
     
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